# Check whether a subset of a vector space is a subspace

$$V = \mathbb{K}^n$$, where $$\mathbb{K}$$ -- field. $$V_1 = \{(x_1,\cdots,x_n)\in V\mid \sum_{i=1}^{n} a_ix_i = 1; a_1,\cdots a_n \in \mathbb{K}\}$$. So, I should check that $$V_1$$ -- subspace.

At this point, I remembered the structure of dual space. So if I have the vector space $$V$$ and the field $$\mathbb{K}$$ (that a $$1$$-dim vec.space under the itself too), and I can construct $$Hom(V,\mathbb{K})$$. This is obviously a vector space, where the sum is determined for functions pointwise (by the sum on images of points). And it is clear that $$Hom(V,\mathbb{K}) = \mathbb{K}^n$$, where $$n = |V|$$ -- power of $$V$$-set. So we can see, that $$Hom(V,\mathbb{K}) = V$$. And if I find the subspace in $$V_{1}^{*} \subset Hom(V,\mathbb{K})$$, which will satisfy the same condition as $$V_1$$ I will proof that $$V_1$$ -- subspace too.

We can check $$(f_1 + f_2)(x) = f_1(x) + f_2(x)$$ and $$(af)(x) = f(ax)$$, for vectors $$x \in V_1$$. If $$f_1(x) = \sum_{i=1}^{n} a_ix_i = 1 = f_2(x)$$ $$\Rightarrow$$ $$(f_1+f_2)(x) = 2$$ -- contradiction.

Is this proof correct?

• @ThomasShelby the $V_1^{*}$ will not be a subspace under this linear condition – Just do it Dec 9 '18 at 4:27
• You haven't defined $V_1^{*}$. – Thomas Shelby Dec 9 '18 at 5:04
• @ThomasShelby The subset of $Hom(V,\mathbb{K})$, which satisfy the same linear condition, like $V_1$. This is should be ok, because $Hom(V,\mathbb{K})$ -- isomorphic vec.space to space $V$ – Just do it Dec 9 '18 at 5:07

Does $$V_1$$ contain the zero vector?